Question

The pressure at the bottom of a cylindrical container with a cross-sectional area of 58.5 cm² and holding a fluid of density 500 kg/m^3 is 115 kPa.

(a) Determine the depth of the fluid.

(b) Determine the pressure at the bottom of the container if an additional 1.80 10^-3 m³ of this fluid is added to the container. (Give your answer to at least 3 significant figures.)

Answer 1

given
cross sectional area, A = 58.5 cm^2
rho = 500 kg/m^3
P = 115 kPa

= 1.15*10^5 pa

a) let h is the depth of the fluid

we know, P = P_atm + rho*g*h

rho*g*h = P - P_atm

h = (P - P_atm)/(rho*g)

= (1.15*10^5 - 1.0134*10^5)/(500*9.8)

= 2.79 m <<<<<<<<<<<-------------------Answer

b) New depth of the cyllinder, H = h + Volume added/A

= 2.79 + 1.80*10^-3/(58.5*10^-4)

= 3.098 m

P = P_atm + rho*g*h

= 1.0134*10^5 + 500*9.8*3.098

= 1.165*10^5 pa (or) 116.5 kPa <<<<<<<<<---------------Answer